Evaluate $\lim_{x\to \infty} \frac{\tan^{2}(\frac{1}{x})}{(\ln(1+\frac{4}{x}))^2}$ $$\lim_{x\to \infty} \frac{\tan^{2}(\frac{1}{x})}{(\ln(1+\frac{4}{x}))^2}$$
I came across this problem and I am having trouble evaluating it. I know that the whole limit will probably be $0$ and that both the numerator and denominator approach $0$.
How do I evaluate it? Using L'Hospital's rule leads to complex expressions, so I don't think that's a good method.
Thank you for the help.
 A: Hint: There are two common limits that are not hard to compute with L'Hopital. They will make computing this limit pretty easy.
$$
\lim_{x\to0}\frac{\tan(x)}{x}=1
$$
and
$$
\lim_{x\to0}\frac{\log(1+x)}{x}=1
$$
A: This is also pretty easy without using the taylor expansion. 
$$L=\lim_{x\to\infty}\frac{\tan^2(1/x)}{\ln^2(1+4/x)}=\left(\lim_{x\to\infty}\frac{\tan(1/x)}{\ln(1+4/x)}\right)^2$$
Let $u=1/x$.
$$\sqrt L=\lim_{u\to0}\frac{\tan(u)}{\ln(1+4u)}=\lim_{u\to0}\frac{\sin(u)}{\cos(u)\ln(1+4u)}$$
$$=\lim_{u\to0}\frac{\sin(u)}{\ln(1+4u)}$$
Now we can apply L'Hospital.
$$\sqrt L=\lim_{u\to 0}\frac{1}{4}\cos(u)(4u+1)=\frac 1 4$$
$$\lim_{x\to\infty}\frac{\tan^2(1/x)}{\ln^2(1+4/x)}=\frac{1}{16}$$
A: Hint: Using the Taylor series we have $$\frac{\tan^{2}\left(\frac{1}{x}\right)}{\log^{2}\left(1+\frac{4}{x}\right)}=\frac{1/x^{2}+O\left(1/x^{6}\right)}{16/x^{2}+O\left(1/x^{4}\right)}.$$
A: Beside the good hint robjohn gave , here again, Taylor expansion make life simple $$A=\frac{\tan^{2}(\frac{1}{x})}{(\ln(1+\frac{4}{x}))^2}$$ Since $x\to \infty$, change $x=\frac 1y$ and consider $$A=\frac{\tan^{2}(y)}{(\ln(1+4y))^2}=\left(\frac{\tan(y)}{\ln(1+4y)}\right)^2$$ Now, close to $y=0$, you know that $\tan(y)\approx y$ and $\log(1+4y)\approx 4y$.
I am sure that you can take from here.
